is a Kan fibration. (Where $\operatorname{Map}$ is the (sSet)-enriched $\operatorname{Hom}$).

Lastly, assume that $A\otimes \Delta^n\tilde{\to}A\otimes \Delta^0=A$ is a weak equivalence for any object $A$ in $C$ and any $n\in \mathbf{N}$. (Here, the tensor $A\otimes K$, where $A$ is in $C$ and $K$ is in $sSet$, is the object of $C$ representing the functor $Map(A,-)^K$).

Let $L\subseteq K$ be an inclusion of simplicial sets. Suppose $\sigma:\Delta^n\hookrightarrow K$ is a nondegenerate simplex of $K$ with all of its faces living in $L$. That is, we can factor the map $\partial\sigma:=\sigma|_{\partial\Delta^n}$ through the inclusion $L\subseteq K$ (in fact, we will assume that the target of this map actually is $L$).

Then for any object $D$ in $C$, the pushout $$D\otimes \Delta^n\coprod_{D\otimes\partial\Delta^n} D\otimes L\cong D\otimes (\Delta^n\coprod_{\partial\Delta^n}L)$$ is a homotopy pushout. Now, the question here is, why is this the case?

The proof I'm reading says that it follows from the line marked (*) above (and the fact that $C$ is left-proper (which follows from the fact that all objects of $C$ are cofibrant)), but it's not clear to me how to apply that hypothesis.

That is, how does the line marked (*) imply anything relevant?

If you'd like to look up the original source, it is Higher Topos Theory Proposition A.3.1.7 (in the appendix).

Edit: It's probable that a few of the hypotheses are unrelated to the actual question. I included everything because I'm not sure what's important.

Here's an easy step, which may or may not be in the right direction: You want to show that tensoring with D preserves cofibrations, or equivalently that its right adjoint preserves trivial fibrations. It's clear that tensoring with D preserves trivial cofibrations, or equivalently that its right adjoint preserves fibrations, because that's what (*) says when A is initial and B is D.
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Tom GoodwillieJul 10 '10 at 17:12

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Well, here's something funny: Not only did my comment not use the assumption that the map $A\otimes\Delta^n\to A\otimes \Delta^0$ is a weak equivalence for any object $A$ -- it shows that that assumption is superfluous, because a one-sided inverse $A\otimes\Delta^0\to A\otimes \Delta^n$ is a trivial cofibration.
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Tom GoodwillieJul 10 '10 at 19:00

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Rereading the question, I find that it looks like you did not write exactly what you meant. I am guessing that $K$ was meant to be the pushout of the simplex and $L$ along the boundary of the simplex -- that is, that $\sigma$ was meant to be the only nondegenerate simplex of $K$ not in $L$. $K$ does not appear in the conclusion.
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Tom GoodwillieJul 11 '10 at 1:37

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I'd just like to comment that the entire question becomes trivial with the correction.
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Harry GindiJul 11 '10 at 7:46

1 Answer
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Assuming I understand the question, isn't the following a counterexample? Let $\cal C$ be sSet with the usual model structure, but with the trivial simplicial enrichment in which the simplicial set $Map(A,X)$ is the discrete (constant) set of sSet morphisms $A\to X$. So $A\otimes K$ is coproduct of $\pi_0(K)$ copies of $A$. Then the inclusion $\partial\Delta^1\to \Delta^1$ induces a map $A\otimes \partial\Delta^1\to A\otimes \Delta^1$ that is not in general a cofibration, and the pushout $A\otimes * \leftarrow A\otimes \partial\Delta^1\to A\otimes \Delta^1$ is not a homotopy pushout.

Or am I making some mistake about the meaning of "enriched, tensored, and cotensored", or about the meaning of "homotopy pushout" in the present context?

well, the example you gave doesn't fit exactly, since both $\partial\Delta^1$ should embed in both $L$ and $\Delta^1$. That wouldn't stop you from coming up with another counterexample, of course.
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Harry GindiJul 11 '10 at 2:17

I didn't interpret "living in $L$" as embedding in $L$. But let's make it $\Delta^1\leftarrow\partial\Delta^1\to\Delta^1$ instead.
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Tom GoodwillieJul 11 '10 at 2:23

Don't take my word for it. Do I have the definitions right? I think that if $C$ is enriched over $V$ then it's said to be tensored if for every $C$-object $A$ the functor $Map(A,-)$ from $C$ to $V$ has a left adjoint; and cotensored if for every $C$-object $B$ the functor $Map(-,B)$ from $C$ to $V^{op}$ has a right adjoint.
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Tom GoodwillieJul 11 '10 at 3:29

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Alright, I've e-mailed Lurie, and he said that the statement of the proposition is incorrect (and hence your counterexample works)!
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Harry GindiJul 11 '10 at 5:07

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Oh, good. I was almost ready to sign myself "enriched, tensored, and bewildered".
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Tom GoodwillieJul 11 '10 at 11:32